Is this solution valid, extraneous, or off-domain? algebra gives \(x=2\) but rational graph has hole at \(x=2\)

Treat an algebraic result as a candidate until it passes the original problem's conditions. A genuine intersection requires both functions to be defined at the input and to produce equal outputs there. On a graph, an open circle marks a missing point rather than a shared point, so domain evidence must take priority over what a simplified equation seems to allow.